Title: Physics 207, Lecture 13, Oct. 15
1Physics 207, Lecture 13, Oct. 15
- Agenda Finish Chapter 10, start Chapter 11
- Chapter 10 Energy
- Potential Energy (gravity, springs)
- Kinetic energy
- Mechanical Energy
- Conservation of Energy
- Start Chapter 11, Work
- Assignment
- HW5 due tonight
- HW6 available today
- Monday, finish reading chapter 11
2Chapter 10 Energy
- Rearranging Newtons Laws gives (Fd vs. ½ mv2
relationship) - -2mg (yf yi ) m (vyf2 - vyi2 )
- or ? ½ m vyi2 mgyi ½ m vyf2 mgyf
- and adding ½ m vxi2 ½ m vzi2 and ½ m
vxf2 ½ m vzf2 -
- ½ m vi2 mgyi ½ m vf2 mgyf
- where vi2 vxi2 vyi2 vzi2
- ½ m v2 terms are referred to as kinetic
energy
3Energy
- If only conservative forces are present, the
total energy (sum of potential, U, and kinetic
energies, K) of a system is conserved.
K ½ mv2 U mgy
Ki Ui Kf Uf
- K and U may change, but E K Umech remains
constant.
Emech is called mechanical energy
4Another example of a conservative system The
simple pendulum.
- Suppose we release a mass m from rest a distance
h1 above its lowest possible point. - What is the maximum speed of the mass and where
does this happen ? - To what height h2 does it rise on the other side ?
5Example The simple pendulum.
- What is the maximum speed of the mass and where
does this happen ? - E K U constant and so K is maximum when U
is a minimum.
y
yh1
y0
6Example The simple pendulum.
- What is the maximum speed of the mass and where
does this happen ? - E K U constant and so K is maximum when U
is a minimum - E mgh1 at top
- E mgh1 ½ mv2 at bottom of the swing
y
yh1
h1
y0
v
7Example The simple pendulum.
- To what height h2 does it rise on the other
side? - E K U constant and so when U is maximum
again (when K 0) it will be at its highest
point. - E mgh1 mgh2 or h1 h2
y
yh1h2
y0
8Lecture 13, Exercise 1Conservation of Mechanical
Energy
A block is shot up a frictionless 40 slope with
initial velocity v. It reaches a height h before
sliding back down. The same block is shot with
the same velocity up a frictionless 20
slope. On this slope, the block reaches height
- 2h
- h
- h/2
- Greater than h, but we cant predict an exact
value. - Less than h, but we cant predict an exact value.
9Lecture 13, ExampleThe Loop-the-Loop again
- To complete the loop the loop, how high do we
have to let the release the car? - Condition for completing the loop the loop
Circular motion at the top of the loop (ac v2 /
R) - Use fact that E U K constant !
Ubmgh
Recall that g is the source of the centripetal
acceleration and N just goes to zero is the
limiting case. Also recall the minimum speed at
the top is
Car has mass m
Umg2R
h ?
R
(A) 2R (B) 3R (C) 5/2 R (D) 23/2 R
10Lecture 13, ExampleThe Loop-the-Loop again
- Use E K U constant
- mgh 0 mg 2R ½ mv2
- mgh mg 2R ½ mgR 5/2 mgR
- h 5/2 R
h ?
R
(A) 2R (B) 3R 5/2 R (D) 23/2 R
11Lecture 13, ExampleSkateboard
- What speed will the skateboarder reach at bottom
of the hill if there is no friction and the
skeateboarder starts at rest? - Assume we can treat the skateboarder as point
- Zero of gravitational potential energy is at
bottom of the hill
m 25 kg
R5 m
R5 m
12Lecture 13, ExampleSkateboard
- What speed will the skateboarder reach at bottom
of the hill if there is no friction and the
skeateboarder starts at rest? - Assume we can treat the skateboarder as point
- Zero of gravitational potential energy is at
bottom of the hill
m 25 kg
- Use E K U constant
- Ebefore Eafter
- 0 mgR ½ mv2 0
- 2gR v2 ? v (2gR)½
- v (2 x 10 x 5)½ 10 m/s
R5 m
R5 m
13Potential Energy, Energy Transfer and Path
- A ball of mass m, initially at rest, is released
and follows three difference paths. All surfaces
are frictionless - The ball is dropped
- The ball slides down a straight incline
- The ball slides down a curved incline
- After traveling a vertical distance h, how do the
three speeds compare?
1
3
2
h
(A) 1 gt 2 gt 3 (B) 3 gt 2 gt 1 (C) 3 2 1
(D) Cant tell
14Lecture 13, Exercise 2 Potential Energy, Energy
Transfer and Path
- A ball of mass m, initially at rest, is released
and follows three difference paths. All surfaces
are frictionless - The ball is dropped
- The ball slides down a straight incline
- The ball slides down a curved incline
- After traveling a vertical distance h, how do the
speeds compare?
- 1 gt 2 gt 3
- 3 gt 2 gt 1
- 3 2 1
- Cant tell
15Potential Energy, Energy Transfer and Path
- A ball of mass m, initially at rest, is released
and follows three difference paths. All surfaces
are frictionless - The ball is dropped
- The ball slides down a straight incline
- The ball slides down a curved incline
- After traveling a vertical distance h, how do the
three speeds compare?
(A) 1 gt 2 gt 3 (B) 3 gt 2 gt 1 (C) 3 2 1
(D) Cant tell
16Elastic vs. Inelastic Collisions
- A collision is said to be elastic when energy as
well as momentum is conserved before and after
the collision.
Kbefore Kafter - Carts colliding with a perfect spring, billiard
balls, etc.
- A collision is said to be inelastic when energy
is not conserved before and after the collision,
but momentum is conserved.
- Kbefore ? Kafter
- Car crashes, collisions where objects stick
together, etc.
17Inelastic collision in 1-D Example 1
- A block of mass M is initially at rest on a
frictionless horizontal surface. A bullet of
mass m is fired at the block with a muzzle
velocity (speed) v. The bullet lodges in the
block, and the block ends up with a speed V. - What is the initial energy of the system ?
- What is the final energy of the system ?
- Is energy conserved?
x
v
V
before
after
18Inelastic collision in 1-D Example 1
- What is the momentum of the bullet with speed v
? -
- What is the initial energy of the system ?
- What is the final energy of the system ?
- Is momentum conserved (yes)?
- Is energy conserved? Examine Ebefore-Eafter
v
No!
V
x
before
after
19Example Fully Elastic Collision
- Suppose I have 2 identical bumper cars.
- One is motionless and the other is approaching it
with velocity v1. If they collide elastically,
what is the final velocity of each car ? - Identical means m1 m2 m
- Initially vGreen v1 and vRed 0
- COM ? mv1 0 mv1f mv2f ? v1 v1f
v2f - COE ? ½ mv12 ½ mv1f2 ½ mv2f2 ? v12 v1f2
v2f2 - v12 (v1f v2f)2 v1f2 2v1fv2f v2f2 ? 2
v1f v2f 0 - Soln 1 v1f 0 and v2f v1 Soln 2 v2f
0 and v1f v1
20Lecture 13, Exercise for homeElastic Collisions
- I have a line of 3 bumper cars all touching. A
fourth car smashes into the others from behind.
Is it possible to satisfy both conservation of
energy and momentum if two cars are moving after
the collision? - All masses are identical, elastic collision.
- (A) Yes (B) No (C) Only in one special
case
v
Before
v1
v2
After?
21Lecture 13, Exercise for homeElastic Collisions
- COM ? mv mv1 mv2 so v v1 v2
- COE ? ½ mv2 ½ mv12 ½ mv22
- v2 (v1 v2)2 v12 v22 ? v1 v2 0
- (A) Yes (B) No (C) Only in one special
case
Before
After?
22Variable force devices Hookes Law Springs
- Springs are everywhere, (probe microscopes, DNA,
an effective interaction between atoms) - In this spring, the magnitude of the force
increases as the spring is further compressed (a
displacement). - Hookes Law,
- Fs - k Dx
- Dx is the amount the spring is stretched or
compressed from it resting position.
Rest or equilibrium position
F
Dx
23Hookes Law Spring
- For a spring we know that Fx -kx.
24Lecture 13, ExampleHookes Law
- Remembering Hookes Law, Fx -k Dx
-
- What are the units for the constant k ?
- (A) (B) (C) (D)
F is in kg m/s2 and dividing by m gives kg/s2 or
N/m
25Lecture 13, Exercise 2Hookes Law
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
26Lecture 10, Exercise 2Hookes Law
8 m
9 m
What is the spring constant k ? SF 0 Fs
mg k Dx - mg Use k mg/Dx 5 N / 0.01 m
Fspring
50 kg
(A) 50 N/m (B) 100 N/m (C) 400 N/m (D) 500 N/m
mg
27F-x relation for a foot arch
Force (N)
Displacement (mm)
28F-x relation for a single DNA molecule
29Measurement technique optical tweezers
30Lecture 13, Oct. 15
- Chapter 10 Energy
- Potential Energy (gravity, springs)
- Kinetic energy
- Mechanical Energy
- Conservation of Energy
- Chapter 11, Work
- Assignment
- HW5 due tonight
- HW6 available today
- Monday, finish Chapter 11